# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x) = x / (3x-1)?

Aug 16, 2016

vertical asymptote at $x = \frac{1}{3}$
horizontal asymptote at $y = \frac{1}{3}$

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: $3 x - 1 = 0 \Rightarrow x = \frac{1}{3} \text{ is the asymptote}$

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by x.

$\frac{\frac{x}{x}}{\frac{3 x}{x} - \frac{1}{x}} = \frac{1}{3 - \frac{1}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{1}{3 - 0}$

$\Rightarrow y = \frac{1}{3} \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1) Hence there are no oblique asymptotes.
graph{(x)/(3x-1) [-10, 10, -5, 5]}